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so here I got a golden rectangle enter image description here Let the point A be $z_0 = 0+0i$

G: $z_1 = 1+i$

J: $z_2 =\phi + (2-\phi)i$

L: $z_3 = 2\phi -2$

N: $z_4 = 1+(2\phi -3 )i$

P: $z_5 = (6-3\phi)+(2-a)i$

etc... I would like to find $\lim_{n\rightarrow\infty}z_n$ but I am struggling trying to find a general expression for $z_n$. I was reading this article: Complex fibonacci numbers

But I don't know how to use the formula $G(n,m)=F_n F_{m-1} +iF_{n+1}F_m$ to approach this problem. Any suggestion?

davidaap
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1 Answers1

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Just look at the picture, each successive square is a factor $\phi-1$ smaller and the vector connecting the corner in the square that interests you to the next one is turning by an angle of $\pi/2$ clockwise. You start with the vector $1+i$, therefore the series becomes

$$(1+i)\sum_{n=0}^{\infty}(\phi-1)^n e^{-ni\pi/2} = \frac{1+i}{1-(\phi-1) e^{-i\pi/2}}=\frac{1+i}{1+(\phi-1)i}.$$

Raskolnikov
  • 16,108